如何從PHP中的DropDown列表中選擇值？問題

Question

我想知道錯誤在此代碼

下面的代碼檢索在 下拉列表

數據庫查詢的成員的名字，但我怎麼知道你選擇誰....我要發送的消息只對選定的形式下拉列表

<?php 
include ("connect.php"); 



    $name = $_POST['sector_list']; 

    echo $name ; 

?> 


    <form method="POST" action="" > 
    <input type="hidden" name="sector" value="sector_list"> 
    <select name="sector_list" class="inputstandard"> 
    <option size ="40" value="default">send to </option> 

    <?php 
     $result = mysql_query('select * from members ') 
     or die (mysql_error()); 

     while ($row = mysql_fetch_assoc($result)) { 

      echo '<option size ="40" value=" '. $row['MemberID'] . '" name="' . $row['MemberName']. '">' . $row['MemberName']. '</option>'; 

     } 

    ?> 

    </select> 
    </form> 

成員，我希望有人能幫助我

Answer 1

這應該做的伎倆。

<?php 
$member_id = intval($_POST['sector_list']); 

if($member_id == 0) { 
    // Default choice was selected 
} 
else { 
    $res = mysql_query("SELECT * FROM members WHERE MemberID = $member_id LIMIT 1"); 
    if(mysql_num_rows($res) == 0) { 
     // Not a valid member 
    } 
    else { 
     // The member is in the database 
    } 
} 
?> 

<form method="post" action=""> 
    <input type="hidden" name="sector" value="sector_list"> 
    <select name="sector_list" class="inputstandard"> 
     <option value="0">send to</option> 
     <?php 
     $result = mysql_query('SELECT * from members') or die(mysql_error()); 

     while ($row = mysql_fetch_assoc($result)) { 
      echo '<option value="' . $row['MemberID'] . '">' . $row['MemberName']. '</option>'; 
     } 
     ?> 
    </select> 
</form> 

爲了得到一個輸入改變，當你選擇一個人試試這個：

<select onchange="document.getElementById('text-input').value = this.value;"> 
<!-- Options here --> 
</select> 
<input type="text" id="text-input">